Separation Process Principles- 2n - Seader &amp; Henley - Solutions Manual

# Therefore company b is dropped 4 exercise 1420

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Unformatted text preview: ) / cm2 - s - cmHg . Convert permeance to American Engineering units of lbmol/ft2-h-psi: 1 barrer/cm = 1 x 10-10 (30.48 cm/ft)2 (76 cmHg/14.696 psi) / (22,400 x 454 cm3 (STP)/lbmol) = 1.70 x 10-10 lbmol/ft2-h-psi Therefore, the permeances are: PM A = 10 × 108 1.70 × 10-10 = 1.70 × 10-2 lbmol / ft 2 - h - psi . 4 1.70 × 10-10 = 3.40 × 10-6 lbmol / ft 2 - h - psi −4 2 × 10 Compute entering and exiting partial pressures. Feed: pA = (1.2)(14.7)(1.25)/250.70 = 0.088 psia pAir = (1.2)(14.7)(249.45)/250.70 = 17.552 psia PM Air = Exercise 14.19 (continued) Analysis: (continued) Retentate: Permeate: pA = (1.2)(14.7)(0.11)/228.00 = 0.0085 psia pAir = (1.2)(14.7)(227.89)/228.00 = 17.6315 psia pA = (5/76)(14.7)(1.14)/22.7 = 0.0486 psia pAir = (5/76)(14.7)(21.56)/22.7 = 0.9185 psia Now check to see if it is even possible to obtain a permeate of 5 mol% acetone (A). For crossflow, the initial acetone content of the local permeate at the feed end must be greater than 5 mol% because the mol% will decrease as the feed moves to the other end. PF = PR = 1.2 atm = 17.64 psia, PP = 5 cmHg = 0.9671 psia, r = PP/PF = 0.9671/17.64 =0.0548 αx Rearranging Eq. (14-36), y = (3) ,where, y and x refer to A and α is for A to air. 1 + (α − 1) x x (α − 1) + 1 − rα x (α − 1) + 1 − 0.0548α = 5000 (4) x (α − 1) + 1 − r x (α − 1) + 1 − 0.0548 Solving Eq. (4), for x = xF = 0.005, α = 19.8872. Substituting this α into Eq. (3), y = 0..0909. Therefore, it is possible to obtain a permeate with a 5 mol% acetone. For this exercise, it is found that α for all values of x can not be assumed constant because the mole fraction of A in the cumulative permeate is very sensitive to the value of α. Therefore, Eq. (4) is used to compute a new α at each step with a new value of x. For crossflow with a binary mixture, we have the following relationships between acetone mole fractions in the permeate and retentate as a function of the cut, θ = nP/nF. x − x R (1 − θ) 0.005 − x R (1 − θ) = From material balance Eq. (1), Example 14.6, y P = F (5) θ θ From Eq. (14-49), where the y and x pertain to A, From Eq. (14-43), α = α * -air A yP = x 1 1− α R 1− θ θ 1 − xR α α −1 xF 1 − xF α α −1 − xR α α −1 =x Combine this equation with Eq. (5) to obtain: 0.005 1− θ = xR + x 1 1− α R 1 − xR α α −1 0.005 0.995 1 1− α R 1− θ θ 1 − xR α α −1 0.005 1 − 0.005 α α −1 − xR (6) α α −1 − xR α α −1 For A, xR begins at 0.005 at the feed end and decreases toward the retentate end. For each step in x, Eq. (4) is used to calculate α. Use Eq. (6) to compute values of θ as xR decreases. Use Eq. (5) to compute the corresponding value of yP. Eq. (3) is used to compute the local value of y. For a differential molar transfer rate, dn, the differential membrane area required is given by Eq. (14-50), which is also given in incremental form for numerical calculation. If we base this calculation on A, α α −1 Exercise 14.19 (continued)...
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